Evaluate $\dfrac c2-3+\dfrac 6d$ when $c=14$ and $d=3$.
Solution: Let's substitute $ c= {14}$ and $ d={3}$ into the expression. $\phantom{=}\dfrac { {c}}2-3+\dfrac6{ d}$ $=\dfrac { {14}}2-3+\dfrac6{ 3}$ $=7-3+2$ $=6$